If in any binomial expansion a, b, c and d be the 6th, 7th, 8th and 9th terms respectively, prove that `(b^2-ac)/(c^2-bd)=(4a)/(3c)`

In the expansion of (x+P)^n ,the 6th ,7th,8th and 9th terms are a,b,c and d resp,show that (b^2-ac)/(c^2-bd)=(4a)/(3c)

If 6th, 7th, 8th and 9th terms of (x+y)^n are a, b,c and d respectively, then prove that : (b^2-ac)/(c^2-bd)=4/3.a/c .

The 4th, 7th and 10 th terms of a GP are a, b, c, respectively. Prove that b^(2)=ac .

The 5th, 8th and 11th terms of a GP are a,b,c respectively. Show that, b^2=ac

If 6th , 7th , 8th, and 9th terms of (x+ y)^n are a,b,c and d respectively , then prove that : (b^2 - ac)/(c^2 -bd) =4/3. a/c

In the binomial expansion of (a-b)^n, n ge 5 , the sum of 5th and 6th terms is zero, then

The 5th, 8th and 11th terms of a GP are a, b, c respectively. Show that b^(2)=ac .

In the binomial expansion (a-b)^n, nge5 the sum of 5th and 6th terms is zero. Then find a/b

If n be a positive integer and if the 3rd, 4th, 5th and 6th terms in the expansion of (x+A)^n, when expanded in ascending powers of x, be a, b, c and d respectively show that, (b^2-ac)/(c^2-bd)=(5a)/(3c)

If the 4th, 7th and 10th terms of a G.P. are a, b, c respectively, then